%252525253Cp%252525253ETwo%2525252520metal%2525252520disks%252525252C%2525252520

Question

%252525253Cp%252525253ETwo%2525252520metal%2525252520disks%252525252C%2525252520one%2525252520with%2525252520radius%2525252520R1%2525252520%252525253D%25252525202.61cm%2525252520and%2525252520mass%2525252520M1%2525252520%252525253D%252525250A0.750kg%2525252520and%2525252520the%2525252520other%2525252520with%2525252520radius%2525252520R2%2525252520%252525253D%25252525205.08cm%2525252520and%2525252520mass%2525252520M2%2525252520%252525253D%25252525201.68kg%252525252C%252525250Aare%2525252520welded%2525252520together%2525252520and%2525252520mounted%2525252520on%2525252520a%2525252520frictionless%2525252520axis%2525252520through%252525250Atheir%2525252520common%2525252520center.%252525253C%252525252Fp%252525253E%252525250A%252525253Cp%252525253EA%252525253A%2525252520What%2525252520is%2525252520the%2525252520total%2525252520moment%2525252520of%2525252520inertia%2525252520of%2525252520the%2525252520two%252525250Adisks%252525253F%252525253C%252525252Fp%252525253E%252525250A%252525253Cp%252525253E%252525253Cspan%2525252520class%252525253D%2525252522c1%2525252522%252525253EB%252525253A%2525252526nbsp%252525253B%252525253C%252525252Fspan%252525253E%252525253Cspan%2525252520class%252525253D%2525252522c2%2525252522%252525253EA%2525252520light%2525252520string%252525250Ais%2525252520wrapped%2525252520around%2525252520the%2525252520edge%2525252520of%2525252520the%2525252520smaller%2525252520disk%252525252C%2525252520and%2525252520a%25252525201.50-kg%252525250Ablock%252525252C%2525252520suspended%2525252520from%2525252520the%2525252520free%2525252520end%2525252520of%2525252520the%2525252520string.%2525252520If%2525252520the%2525252520block%2525252520is%252525250Areleased%2525252520from%2525252520rest%2525252520at%2525252520a%2525252520distance%2525252520of%25252525201.98m%2525252520above%2525252520the%2525252520floor%252525252C%2525252520what%2525252520is%252525250Aits%2525252520speed%2525252520just%2525252520before%2525252520it%2525252520strikes%2525252520the%2525252520floor%252525253F%252525253C%252525252Fspan%252525253E%252525253C%252525252Fp%252525253E%252525250A%252525253Cp%252525253E%252525253Cspan%2525252520class%252525253D%2525252522c2%2525252522%252525253EC%252525253A%2525252526nbsp%252525253BRepeat%2525252520the%2525252520calculation%2525252520of%2525252520part%2525252520B%252525252C%2525252520this%252525250Atime%2525252520with%2525252520the%2525252520string%2525252520wrapped%2525252520around%2525252520the%2525252520edge%2525252520of%2525252520the%2525252520larger%252525250Adisk.%2525252526nbsp%252525253B%252525253C%252525252Fspan%252525253E%252525253C%252525252Fp%252525253E%252525250A%252525253Cp%252525253E%252525253Cimg%2525252520src%252525253D%252525250A%2525252522http%252525253A%252525252F%252525252Fsession.masteringphysics.com%252525252FproblemAsset%252525252F1260432%252525252F3%252525252FYF-09-37.jpg%2525252522%252525250Aalt%252525253D%2525252522no%2525252520title%2525252520provided%2525252522%2525252520%252525252F%252525253E%252525253C%252525252Fp%252525253E%252525250A

Explanation / Answer

Weight of combination W = 23.6 N I1 (about its edge) = 3/2 M1R12 = 6.77 * 10E-4 I2 (about point at distance R1 from itscenter) = 1/2 M2 R22 +M2 R12 I2 = 1.90 * 10E-3 + 8.91 * 10E-4 = 2.78 *10E-3 I = I1 + I2 = 3.46 * 10E-3 Taking the torque about point of contact of string withR1    L = W R1 L = I ? = I a / R1   and  a = W R12 / I = 23.6 * 5.57 * 10E-4 / 3.46 *10E-3 = 3.80 m/s^2 v^2 = 2 a h = 2 * 3.80 * 2.04 = 15.5   and v = 3.94m/s For the second part you need to calculate the moment inertiaof the combination about the edge of R2 and calculate thecorresponding torque- using the moments of inertia about the point of contact withthe stringeliminates the need to account for the tension in the string. L = I ? = I a / R1   and  a = W R12 / I = 23.6 * 5.57 * 10E-4 / 3.46 *10E-3 = 3.80 m/s^2 v^2 = 2 a h = 2 * 3.80 * 2.04 = 15.5   and v = 3.94m/s For the second part you need to calculate the moment inertiaof the combination about the edge of R2 and calculate thecorresponding torque- using the moments of inertia about the point of contact withthe stringeliminates the need to account for the tension in the string.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---